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24. Swap Nodes in Pairs

时间:2015-05-14 08:34:25      阅读:75      评论:0      收藏:0      [点我收藏+]

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

实现题,注意边界情况即可

/**
* Definition for singly-linked list.
* public class ListNode {
*    int val;
*    ListNode next;
*    ListNode(int x) { val = x; }
* }
*/
public class Solution {
  public ListNode swapPairs(ListNode head) {
    ListNode newHead = new ListNode(0);
    newHead.next = head;
    head = newHead;
    while (head!= null && head.next != null && head.next.next != null) {
      ListNode next = head.next;
      head.next = head.next.next;
      next.next = head.next.next;
      head.next.next = next;
      head = head.next.next;
    }
    return newHead.next;
  }
}

24. Swap Nodes in Pairs

原文:http://www.cnblogs.com/shini/p/4502322.html

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